Answer to Question #127626 in Physics for Abiona

Question

The temperature of an object is given byT(t) = 280 + 1.5t^2 e^(?0.12t) where t is measured in minutes and T is measured in Kelvins (abbreviated K). Answer the following questions. Round all answers to two decimal places. Include correct units.

(a) Find the rate of change of the rate of change of temperature at the instant t = 20 minutes.

Answer 1

$$T(t)=280+1.5t^2 \exp{(0.12t)}$$

The rate of change of temperature

$$\frac{dT}{dt}=3.00t \exp{(0.12t)}+0.18t^2 \exp{(0.12t)}$$

The rate of change of the rate of change of temperature

$$\frac{d^2T}{dt^2}=3.00 \exp{(0.12t)}\\ +0.72t^2 \exp{(0.12t)}+0.0216t^2 \exp{(0.12t)}$$

At the instant $t=20$ minutes we get

$$\frac{d^2T}{dt^2}=287.04\:\rm K/min^2$$